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The nearest star to our solar system is 4.29 light years away. How much is this distance in terms of parsecs? How much parallax would this star (named AlphaCentauri) show when viewed from two locations of the Earth six months apart in its orbit around the Sun ?


Distance space of space the space star space from space the space solar space system space equals space 4.29 space ly

1 space light space year space equals space Speed space of space light space cross times space 1 space year

space space space space space space space space space space space space space space space space space space space space space equals space 3 space cross times space 10 to the power of 8 space cross times space 365 space cross times space 24 space cross times space 60 space cross times space 60 space

space space space space space space space space space space space space space space space space space space space space equals space 94608 space cross times space 10 to the power of 11 space straight m space

therefore space 4.29 space ly space equals space 405868.32 space cross times space 10 to the power of 11 space straight m
space space space space space space space space space space space space space space space space space space space space space space space space space left square bracket space 1 space parsec space equals space 3.08 cross times 10 to the power of 16 space straight m right square bracket space

therefore space 4.29 space ly space equals fraction numerator 405868.32 space cross times space 10 to the power of 11 space straight m over denominator space 3.08 cross times 10 to the power of 16 space straight m end fraction space equals space 1.32 space parsec

Now comma space using space the space formula comma space

straight theta space equals space straight d over straight D space
where comma space

straight d space equals space diameter space of space Earth apostrophe straight s space orbit space equals space 3 cross times 10 to the power of 11 space straight m

straight D space equals space Distance space of space star space from space the space Earth space equals 405868.32 space cross times space 10 to the power of 11 space end exponent straight m

therefore space straight theta space equals space fraction numerator 3 cross times 10 to the power of 11 space straight m over denominator 405868.32 space cross times space 10 to the power of 11 space end exponent straight m end fraction

space space space space space space space space equals space 7.39 space cross times space 10 to the power of negative 6 space end exponent rad space

But comma space 1 space sec space equals space 4.85 space cross times space 10 to the power of – 6 end exponent space rad

therefore space space 7.39 space cross times space 10 to the power of negative 6 space end exponent rad space space equals space fraction numerator 7.39 space cross times space 10 to the power of negative 6 space end exponent over denominator space 4.85 space cross times space 10 to the power of – 6 end exponent space end fraction space equals space 1.52 space "
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The principle of ‘parallax’ in section 2.3.1 is used in the determination of distances of very distant stars. The baseline AB is the line joining the Earth’s two locations six months apart in its orbit around the Sun. That is, the baseline is about the diameter of the Earth’s orbit 3 × 1011m. However, even the nearest stars are so distant that with such a long baseline, they show parallax only of the order of 1” (second) of arc or so. A parsec is a convenient unit of length on the astronomical scale. It is the distance of an object that will show a parallax of 1” (second) of arc from opposite ends of a baseline equal to the distance from the Earth to the Sun. How much is a parsec in terms of metres ?

Diameter of Earth’s orbit = 3 × 1011 m

Radius of Earth’s orbit, r = 1.5 × 1011 m

Let the distance parallax angle be 1" = 4.847 × 10–6 rad 

Let the distance of the star be D.

Parsec is defined as the distance at which the average radius of the Earth’s orbit subtends an angle of 1"

Therefore comma space

we space have space straight theta space equals space straight r over straight D space

rightwards double arrow space straight D space equals space straight r over straight theta space equals space fraction numerator 1.5 cross times space 10 to the power of 11 over denominator 4.847 space cross times space 10 to the power of negative 6 end exponent end fraction space

space space space space space space space space space equals space space 0.309 space cross times space 10 to the power of negative 6 space end exponent almost equal to space 3.09 space cross times space 10 to the power of 16 space straight m space

space

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A man walking briskly in rain with speed v must slant his umbrella forward making an angle  with the vertical. A student derives the following relation between θ and v: tan θ = v and checks that the relation has a correct limit : as v →0, θ → 0 as expected. (We are assuming there is no strong wind and that the rain falls vertically for a stationary man.) Do you think this relation can be correct?  If not, guess the correct relation. 


The relation tan straight theta = v is wrong because the dimensions on both the sides of the equation are not equal.

Dimension of right hand side is [MoL1T-1

Dimension of Left hand side quantity is [MoLoTo

Hence, the equation is wrong. 

To make the relation dimensionally correct, R.H.S should be divided by the speed of the rainfall. 

That is, 

tan space theta space equals space fraction numerator v over denominator v apostrophe end fraction, is a dimensionally correct relation. 
521 Views

Precise measurements of physical quantities are a need of science. For example, to ascertain the speed of an aircraft, one must have an accurate method to find its positions at closely separated instants of time. This was the actual motivation behind the discovery of radar in World War II. Think of different examples in modern science
where precise measurements of length, time, mass etc. are needed. Also, wherever you can, give a quantitative idea of the precision needed.


It is indeed very true that precise measurements of physical quantities are essential for the development of science.

For example, ultra-shot laser pulses (time interval ∼ 10
–15 s) are used to measure time intervals in several physical and chemical processes.
X-ray spectroscopy is used to determine the inter atomic separation or inter-planer spacing. 

The development of mass spectrometer makes it possible to measure the mass of atoms precisely.
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Explain this common observation clearly : If you look out of the window of a fast moving train, the nearby trees, houses etc. seem to move rapidly in a direction opposite to the train’s motion, but the distant objects (hill tops, the Moon, the stars etc.) seem to be stationary. (In fact, since you are aware that you are moving, these distant objects seem to move with you).


Line of sight is defined as an imaginary line joining an object and an observer's eye. When we observe nearby stationary objects such as trees, houses, etc. while sitting in a moving train, they appear to move rapidly in the opposite direction because the line of sight changes very rapidly.

On the other hand, distant objects such as trees, stars, etc. appear stationary because of the large distance.

As a result, the line of sight does not change its direction rapidly.
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